package  main.java.leetcode.editor.cn;
//2022-03-29 17:42:28
//编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。 
//
// 不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。 
//
// 
//
// 示例 1： 
//
// 
//输入：s = ["h","e","l","l","o"]
//输出：["o","l","l","e","h"]
// 
//
// 示例 2： 
//
// 
//输入：s = ["H","a","n","n","a","h"]
//输出：["h","a","n","n","a","H"] 
//
// 
//
// 提示： 
//
// 
// 1 <= s.length <= 105 
// s[i] 都是 ASCII 码表中的可打印字符 
// 
// Related Topics 递归 双指针 字符串 
// 👍 552 👎 0

class ReverseString {
    public static void main(String[] args) {
        //创建该题目的对象方便调用
        Solution solution = new ReverseString().new Solution();
        char[] s = {'h','e','l','l','o'};
        solution.reverseString(s);
        new Short().mai();
    }
    public static class Short {
        public  void mai() {
            StringBuffer s = new StringBuffer("Hello");
            if ((s.length() > 5) && (s.append("there").equals("False")));
            System.out.println("value is " + s);
        }
    }
    //leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public void reverseString(char[] s) {
        int slowIndex = 0;
        int fastIndex = s.length-1;
//        while (slowIndex<=fastIndex){
//            char key = s[slowIndex];
//            s[slowIndex] = s[fastIndex];
//            s[fastIndex] = key;
//            slowIndex++;
//            fastIndex--;
//        }
        swap(s,slowIndex,fastIndex);
    }

    public void swap(char[] s,int slowIndex,int fastIndex){
        char key = s[slowIndex];
        s[slowIndex] = s[fastIndex];
        s[fastIndex] = key;
        slowIndex++;
        fastIndex--;
        if(fastIndex>=slowIndex){
            swap(s,slowIndex,fastIndex);
        }
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}
